GATE CS 2026 Set 1 Q49 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the total time for the data transfer, we calculate the seek time, average rotational latency, and transfer time.

1. Rotational Time Calculations:

• Rotational speed = $15000$ rpm = $\frac{15000}{60} = 250$ rotations per second.
• Time for one full rotation ($T_{\text{rot}}$) = $\frac{1}{250}\text{ s} = 4\text{ ms}$.
• Average rotational latency ($T_{\text{latency}}$) = $\frac{T_{\text{rot}}}{2} = 2\text{ ms}$.
• Transfer time for one sector ($T_{\text{transfer}}$) = $\frac{4\text{ ms}}{400\text{ sectors}} = 0.01\text{ ms}$.

2. Seek Time:

The head moves in the order: Track $0 \to 5 \to 12 \to 7$.

• Track $0 \to 5$: $5 \times 1\text{ ms} = 5\text{ ms}$
• Track $5 \to 12$: $7 \times 1\text{ ms} = 7\text{ ms}$
• Track $12 \to 7$: $5 \times 1\text{ ms} = 5\text{ ms}$
• Total seek time = $5 + 7 + 5 = 17\text{ ms}$.

3. Sector Access Time:

Since the $10$ sectors on each of the $3$ tracks are randomly located, each sector is accessed individually.

• Total number of sectors to transfer = $3 \times 10 = 30$ sectors.
• Time to access one random sector = $T_{\text{latency}} + T_{\text{transfer}} = 2 + 0.01 = 2.01\text{ ms}$.
• Total sector access time = $30 \times 2.01\text{ ms} = 60.3\text{ ms}$.

4. Total Time:

$$\text{Total Time} = \text{Total Seek Time} + \text{Total Sector Access Time} = 17\text{ ms} + 60.3\text{ ms} = 77.3\text{ ms}$$